Solve sqrt{x-1}+sqrt{x+1}=4

$$\sqrt { x - 1 } + \sqrt { x + 1 } = 4$$

Solve for x

$$x = \frac{65}{16} = 4\frac{1}{16} = 4.0625$$

Solution Steps

Subtract \sqrt{x+1} from both sides of the equation.

$$\sqrt{x-1}=4-\sqrt{x+1}$$

Square both sides of the equation.

$$\left(\sqrt{x-1}\right)^{2}=\left(4-\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x-1} to the power of 2 and get x-1.

$$x-1=\left(4-\sqrt{x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{x+1}\right)^{2}.

$$x-1=16-8\sqrt{x+1}+\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$x-1=16-8\sqrt{x+1}+x+1$$

Add 16 and 1 to get 17.

$$x-1=17-8\sqrt{x+1}+x$$

Add 8\sqrt{x+1} to both sides.

$$x-1+8\sqrt{x+1}=17+x$$

Subtract x from both sides.

$$x-1+8\sqrt{x+1}-x=17$$

Combine x and -x to get 0.

$$-1+8\sqrt{x+1}=17$$

Add 1 to both sides.

$$8\sqrt{x+1}=17+1$$

Add 17 and 1 to get 18.

$$8\sqrt{x+1}=18$$

Divide both sides by 8.

$$\sqrt{x+1}=\frac{18}{8}$$

Reduce the fraction \frac{18}{8} to lowest terms by extracting and canceling out 2.

$$\sqrt{x+1}=\frac{9}{4}$$

Square both sides of the equation.

$$x+1=\frac{81}{16}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{81}{16}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{81}{16}-1$$

Subtract 1 from \frac{81}{16}.

$$x=\frac{65}{16}$$

Substitute \frac{65}{16} for x in the equation \sqrt{x-1}+\sqrt{x+1}=4.

$$\sqrt{\frac{65}{16}-1}+\sqrt{\frac{65}{16}+1}=4$$

Simplify. The value x=\frac{65}{16} satisfies the equation.

$$4=4$$

Equation \sqrt{x-1}=-\sqrt{x+1}+4 has a unique solution.

$$x=\frac{65}{16}$$